Let $A = \{n : n \in \mathbb{N}, n \text{ is a 3-digit number}\}$, $B = \{9k + 2 : k \in \mathbb{N}\}$, and…

$B$ and $C$ will contain three digit numbers of the form $9k+2$ and $9k+l$ respectively. We need to find sum of all elements in the set $B\cup C$ effectively.

Now, $S\left(B\cup C\right)=S\left(B\right)+S\left(C\right)-S\left(B\cap C\right)$ where $S\left(k\right)$ denotes sum of elements of set $k.$

Also, $B=\left\{101,110,\dots \dots ,992\right\}$

$\therefore S\left(B\right)=\frac{100}{2}\left(101+992\right)=54650$

Case-I: If $l=2$

then $B\cap C=B$

$\therefore S\left(B\cup C\right)=S\left(B\right)$

which is not possible as given sum is

$274\times 400=109600.$

Case-II If $l\ne 2$

then $B\cap C=\varphi$

$\therefore S\left(B\cup C\right)=S\left(B\right)+S\left(C\right)=400\times 274$

$\Rightarrow 54650+\sum _{k=1}^{110}9k+l=109600$

$\Rightarrow 9\sum _{k=11}^{110}k+\sum _{k=11}^{110}l=54950$

$\Rightarrow 9\left(\frac{100}{2}\left(11+110\right)\right)+l\left(100\right)=54950$

$\Rightarrow 54450+100l=54950$

$\Rightarrow l=5$